# Boundary Edge extremities

Hello all,

I would like to know how to have access to edge extremities (x1,y1) and (x2,y2) when we calculate for example:

`````` int1d(ThK)(u*((x - x1)*(x2 - x1) + (y - y1)*(y2 - y1))/( (x2 - x1)^2 + (y2 – y1)^2 ))
``````

The problem reads how to find the two vertices of the current edge.

Thank you in advance for your help,

Best regards,

Loïc,

Idea,

1. ( (x2 - x1)^2 + (y2 – y1)^2 )) = lenEdge^2
2. (x-x1)/(x2-x1) == (y-y1)/(y2=y1) == barycentric coordinate en edge…
``````load "Element_PkEdge"
mesh Th=square(2,1);
func Tg = [ -N.y,N.x];//  Tangente to edge
fespace Wh(Th,P1edge);
Wh b,bb;
//  int_E (b) = 0.5 ; on each e edge
//   (grad(b).T) =  1 ; on each e edge T the tangent to E
verbosity = 10;
plot(b,wait=1);

``````

`b` give le barycentric on edge if I a make no mistake.

Hello @frederichecht ,

Thank you for you answer,
Your method seems to work in this case, but I don’t understand what “ComputeB(b,bb)”
does. Could you give me more insights about this calculation ?

In more general case, if I do not consider the barycentric coordinate on edge, but I consider

int1d(ThK)(u*P(x,y, [x1,y1], [x2, y2]))

where P is a polynomial which depends on the edge extremities [x1,y1], [x2, y2]). Do you have an idea how to have access at this extremities ? Or may be it will be possible to compute P in the say way that in the method you have proposed.

Best regards,

Loïc,

The method compute a barycentric coordinate on the edge
so b(x,y) is a such that X = b(X)*X2 + (1-b(X))*X1
so you can easily make the change of variable with a other polynom.

Otherwise you can use Pkedge with k = 0 to 5 to set a polynom on each edge

the degre of freedom are the Quadrature point of the Gauss Legendre formulae on [0,1] with k point

Thank you for your clear explanation @frederichecht. It is exactely what I was looking for.

However, I just don’t understand where the formula:

``````intalledges(Th,qforder=1)(b*bb+grad(b)'*grad(bb)) -  intalledges(Th,qforder=1)(0.5*bb+Tg'*grad(bb))
``````

comes from. Could you explain it briefly ?

Best regards,

Loïc,

This simple the barycentric coordinate verif

b( middle) = 0.5
and grad(b) . T = 1

qforder = 1 => one quadrature point at middle of E
=> OK